UVA Wall

这个题也很简单。  就是求凸包。  然后 求外围凸包的和就行。 问题就出在 没一个点的地方的处理。

因为是求最小的距离。那么在点的地方。 外围就是要求最小就是圆弧。  对于 所有的点的圆弧 加起来 正好是一个完整的圆（因为要围一圈）。

所以 最小距离 就是  凸包的距离  +   圆的周长。

#include <cstdio>#include <algorithm>#include <iostream>#include <cstring>#include <cmath>#include <cstdlib>#include <string>#include <map>#include <vector>#include <set>#include <queue>#include <stack>#include <cctype>using namespace std;#define ll long longtypedef unsigned long long ull;#define maxn 1001#define INF 1<<30struct Point{    double x,y;    Point(double x = 0, double y = 0):x(x),y(y) {}};typedef Point Vector ;Vector operator + (Vector A, Vector B){ return Vector(A.x + B.x, A.y + B.y);}Vector operator - (Vector A, Point B){return Vector(A.x - B.x, A.y - B.y);}Vector operator * (Vector A, double p){return Vector(A.x * p, A.y * p);}Vector operator / (Vector A, double p){return Vector(A.x / p, A.y / p);}bool operator < (const Point & a, const Point & b){    return a.x < b.x || (a.x == b.x && a.y < b.y);}const double eps = 1e-10;int dcmp(double x){    if(fabs(x) < eps) return 0;    else return x < 0 ? -1 : 1;}bool operator == (const Point & a, const Point & b){    return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;}double Dot(Vector A, Vector B){ return A.x * B.x + A.y * B.y;} // 点积double Length(Vector A) { return sqrt(Dot(A, A));}             //向量长度double Angle(Vector A, Vector B){ return acos(Dot(A, B)/Length(A)/Length(B));}                                                                //夹角double Cross(Vector A, Vector B){return A.x*B.y - A.y * B.x;}   //叉积（面积两倍）double Area2(Point A, Point B, Point C){return Cross(B-A,C-A);} // 面积两倍Vector Rotate(Vector A, double rad){return Vector(A.x*cos(rad) - A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad));}// 旋转后的直线</span>Vector Normal(Vector A){    double L = Length(A);    return Vector(-A.y/L, A.x/L);}Point GetLineIntersection(Point P, Point v, Point Q, Point w){  //求直线 pv 与qw的交点    Vector u = P-Q;    double t = Cross(w, u) / Cross(v, w);    return P+v*t;}double DistanceToline(Point P, Point A, Point B){                //点到直线的距离    Vector v1 = B - A,v2 = P - A;    return fabs(Cross(v1,v2))/Length(v1);}double DistanceTpsegment(Point P, Point A, Point B){             //点到线段的距离    if(A == B) return Length(P-A);    Vector v1 = B - A, v2 = P - A, v3 = P - B;    if(dcmp(Dot(v1, v2)) < 0) return Length(v2);    else if(dcmp(Dot(v1,v3)) > 0) return Length(v3);    else return fabs(Cross(v1, v2)) / Length(v1);}Point GetLinePrijection(Point P, Point A, Point B){             // 点在直线上的投影    Vector v = B-A;    return A+v*(Dot(v, P-A)/Dot(v,v));}bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2){ // 线段相交判定（不包括在端点处的情况）    double c1 = Cross(a2 - a1,b1-a1) ,c2 = Cross(a2 - a1,b2-a1),    c3 = Cross(b2 - b1, a1-b1), c4 = Cross(b2 - b1, a2 - b1);    return dcmp(c1)*dcmp(c2) < 0 && dcmp(c3)*dcmp(c4) < 0;}bool OnSegment(Point p, Point a1, Point a2){                              //线段在端点处是否可能相交    return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;}double ConvexPolygonArea(Point * p,int n){                         // 多边形的有向面积    double area = 0;    for(int i = 1; i < n-1; i++)        area += Cross(p[i] - p[0],p[i+1] - p[0]);    return area/2;}int ConvexHull(Point *p, int n, Point * ch){    sort(p, p+n);    int m = 0;    for(int i = 0; i < n; i++){        while(m > 1 && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;        ch[m++] = p[i];    }    int k = m;    for(int i = n-2; i >= 0; i--){        while(m > k && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;        ch[m++] = p[i];    }    if(n > 1) m --;    return m;}int main (){    int kase;    int counts = 0;    scanf("%d",&kase);    while(kase--){        if(counts)        printf("\n");        counts++;        int n;        double r;        Point p[maxn];        Point af[maxn];        scanf("%d%lf",&n,&r);        for(int i = 0; i < n; i++)            scanf("%lf%lf",&p[i].x,&p[i].y);        int m = ConvexHull(p, n, af);        double sum = 0;        for(int i = 0; i < m; i++)            sum += Length(af[i] - af[(i+1)%m]);        double pi = acos(-1);        sum += pi*r*2;        printf("%.0lf\n",sum);    }    return 0;}